Compute the volume of the solid region bounded below by the cone z = \sqrt {x^2 + y^2} and above...
Question:
Compute the volume of the solid region bounded below by the cone {eq}z = \sqrt {x^2 + y^2} {/eq} and above by the sphere {eq}x^2 + y^2 + z^2 = 4. {/eq}
Volume:
A region's volume, which is bounded below by a cone {eq}\;z = f\left( {x,y} \right)\; {/eq}and above by a sphere {eq}\;g\left( {x,y,z} \right), {/eq} can be evaluated by triple integration. Additionally, cylindrical coordinates are beneficial to calculate the volume of the assigned region. For the cylindrical coordinates:{eq}\;x = r\cos \theta ,y = r\sin \theta ,z = z.\; {/eq}The formula to obtain volume is represented as: {eq}\;\int {\int {\int_D {dV} } } = \int {\int {\int_D {rdzdrd\theta } } }. {/eq}
Answer and Explanation: 1
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- The equation of the cone is {eq}\;z = \sqrt {{x^2} + {y^2}} . {/eq}
- The equation of the sphere is {eq}\;{x^2} + {y^2} + {z^2} = 4. {/eq}
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